Dado un árbol binario , la tarea es comprimir todos los Nodes en la misma línea vertical en un solo Node de tal manera que si el conteo de bits establecidos de todos los Nodes en una línea vertical en cualquier posición es mayor que el conteo de bits claros en esa posición, entonces se establece el bit del Node único en esa posición.
Ejemplos:
Aporte:
1 \ 2 / 1 \ 3Salida: 1 2
Explicación:
1 y 1 están a la misma distancia vertical, cuenta del bit establecido en la posición 0 = 2 y cuenta del bit borrado = 0. Por lo tanto, se establece el bit 0 del Node resultante.
2 y 3 están a la misma distancia vertical, el recuento de bits establecidos en la pos. 0 = 1 y el recuento de bits borrados = 1. Por lo tanto, se establece el bit 0 del Node resultante no se establece.
2 y 3 están a la misma distancia vertical, el recuento de bits establecidos en la primera posición = 2 y el recuento de bits borrados = 0. Por lo tanto, se establece el primer bit del Node resultante.
Aporte:
1 / \ 3 2 / \ / \ 1 4 1 2Salida: 1 3 5 2 2
Enfoque: La idea es atravesar el árbol y mantener un registro de la distancia horizontal desde el Node raíz para cada Node visitado. A continuación se muestran los pasos:
- Se puede utilizar un mapa para almacenar la distancia horizontal desde el Node raíz como clave y los valores de los Nodes como valores .
- Después de almacenar los valores en el mapa, verifique el número de bits establecidos y no establecidos para cada posición y calcule los valores en consecuencia.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Structure of a node of the tree
class Node {
public:
int val;
Node *left, *right;
Node(int val)
{
this->val = val;
left = right = NULL;
}
};
// Function to compress all the nodes on the same vertical
// line
void evalComp(vector<int>& arr)
{
// Stores node by compressing all nodes on the current
// vertical line
int ans = 0;
// Check if i-th bit of current bit set or not
int getBit = 1;
// Iterate over the range [0, 31]
for (int i = 0; i < 32; i++) {
// Stores count of set bits at i-th positions
int S = 0;
// Stores count of clear bits at i-th positions
int NS = 0;
// Traverse the array
for (auto j : arr) {
// If i-th bit of current element is set
if (getBit & j)
S++; // Update S
else
NS++; // Update NS
}
// If count of set bits at i-th position is greater
// than count of clear bits
if (S > NS)
// Update ans
ans += pow(2, i);
// Update getBit
getBit <<= 1;
}
cout << ans << " ";
}
// Function to traverse the tree and map all the nodes of
// same vertical line to vertical distance
void Trav(Node* root, int hd, map<int, vector<int> >& mp)
{
if (!root)
return;
// Storing the values in the map
mp[hd].push_back(root->val);
// Recursive calls on left and right subtree
Trav(root->left, hd - 1, mp);
Trav(root->right, hd + 1, mp);
}
// Function to compress all the nodes on the same vertical
// line with a single node that satisfies the condition
void compressTree(Node* root)
{
// Map all the nodes on the same vertical line
map<int, vector<int> > mp;
Trav(root, 0, mp);
// Getting the range of horizontal distances
int lower, upper;
for (auto i : mp) {
lower = min(lower, i.first);
upper = max(upper, i.first);
}
for (int i = lower; i <= upper; i++)
evalComp(mp[i]);
}
// Driver Code
int main()
{
Node* root = new Node(5);
root->left = new Node(3);
root->right = new Node(2);
root->left->left = new Node(1);
root->left->right = new Node(4);
root->right->left = new Node(1);
root->right->right = new Node(2);
// Function Call
compressTree(root);
return 0;
}
// This code is contributed by Tapesh(tapeshdua420)
Java
// Java program for the above approach
import java.util.*;
class GFG {
// Structure of a node of the tree
static class Node {
int val;
Node left, right;
Node(int val)
{
this.val = val;
left = right = null;
}
}
// Driver Code
public static void main(String[] args)
{
Node root = new Node(5);
root.left = new Node(3);
root.right = new Node(2);
root.left.left = new Node(1);
root.left.right = new Node(4);
root.right.left = new Node(1);
root.right.right = new Node(2);
compressTree(root);
}
// Function to compress all the nodes on the same
// vertical line
public static void evalComp(ArrayList<Integer> arr)
{
// Stores node by compressing all nodes on the
// currentvertical line
int ans = 0;
// Check if i-th bit of current bit set or not
int getBit = 1;
// Iterate over the range [0, 31]
for (int i = 0; i < 32; i++) {
// Stores count of set bits at i-th positions
int S = 0;
// Stores count of clear bits at i-th positions
int NS = 0;
// Traverse the array
for (int j : arr) {
// If i-th bit of current element is set
if ((getBit & j) != 0)
S++; // Update S
else
NS++; // Update NS
}
// If count of set bits at i-th position is
// greater
// than count of clear bits
if (S > NS)
// Update ans
ans += Math.pow(2, i);
// Update getBit
getBit <<= 1;
}
System.out.print(ans + " ");
}
// Function to traverse the tree and map all the nodes
// of same vertical line to vertical distance
public static void
Trav(Node root, int hd,
HashMap<Integer, ArrayList<Integer> > mp)
{
if (root == null)
return;
// Storing the values in the map
mp.putIfAbsent(hd, new ArrayList<>());
mp.get(hd).add(root.val);
// Recursive calls on left and right subtree
Trav(root.left, hd - 1, mp);
Trav(root.right, hd + 1, mp);
}
// Function to compress all the nodes on the same
// vertical
// line with a single node that satisfies the condition
public static void compressTree(Node root)
{
// Map all the nodes on the same vertical line
HashMap<Integer, ArrayList<Integer> > mp
= new HashMap<>();
Trav(root, 0, mp);
// Getting the range of horizontal distances
int lower = Integer.MAX_VALUE, upper
= Integer.MIN_VALUE;
for (Map.Entry<Integer, ArrayList<Integer> > i :
mp.entrySet()) {
lower = Math.min(lower, i.getKey());
upper = Math.max(upper, i.getKey());
}
for (int i = lower; i <= upper; i++)
evalComp(mp.get(i));
}
}
// This code is contributed by Tapesh(tapeshdua420)
Python3
# Python3 program for the above approach
# Structure of a node
# of th tree
class TreeNode:
def __init__(self, val ='', left = None, right = None):
self.val = val
self.left = left
self.right = right
# Function to compress all the nodes
# on the same vertical line
def evalComp(arr):
# Stores node by compressing all
# nodes on the current vertical line
ans = 0
# Check if i-th bit of current bit
# set or not
getBit = 1
# Iterate over the range [0, 31]
for i in range(32):
# Stores count of set bits
# at i-th positions
S = 0
# Stores count of clear bits
# at i-th positions
NS = 0
# Traverse the array
for j in arr:
# If i-th bit of current element
# is set
if getBit & j:
# Update S
S += 1
else:
# Update NS
NS += 1
# If count of set bits at i-th position
# is greater than count of clear bits
if S > NS:
# Update ans
ans += 2**i
# Update getBit
getBit <<= 1
print(ans, end = " ")
# Function to compress all the nodes on
# the same vertical line with a single node
# that satisfies the condition
def compressTree(root):
# Map all the nodes on the same vertical line
mp = {}
# Function to traverse the tree and map
# all the nodes of same vertical line
# to vertical distance
def Trav(root, hd):
if not root:
return
# Storing the values in the map
if hd not in mp:
mp[hd] = [root.val]
else:
mp[hd].append(root.val)
# Recursive calls on left and right subtree
Trav(root.left, hd-1)
Trav(root.right, hd + 1)
Trav(root, 0)
# Getting the range of
# horizontal distances
lower = min(mp.keys())
upper = max(mp.keys())
for i in range(lower, upper + 1):
evalComp(mp[i])
# Driver Code
if __name__ == '__main__':
root = TreeNode(5)
root.left = TreeNode(3)
root.right = TreeNode(2)
root.left.left = TreeNode(1)
root.left.right = TreeNode(4)
root.right.left = TreeNode(1)
root.right.right = TreeNode(2)
# Function Call
compressTree(root)
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
// Structure of a node of the tree
class Node {
public int val;
public Node left;
public Node right;
public Node(int data)
{
val = data;
left = right = null;
}
}
class Program {
// Driver Code
static void Main(string[] args)
{
Node root = new Node(5);
root.left = new Node(3);
root.right = new Node(2);
root.left.left = new Node(1);
root.left.right = new Node(4);
root.right.left = new Node(1);
root.right.right = new Node(2);
compressTree(root);
}
// Function to compress all the nodes on the same
// vertical line
public static void evalComp(List<int> arr)
{
// Stores node by compressing all nodes on the
// currentvertical line
int ans = 0;
// Check if i-th bit of current bit set or not
int getBit = 1;
// Iterate over the range [0, 31]
for (int i = 0; i < 32; i++) {
// Stores count of set bits at i-th positions
int S = 0;
// Stores count of clear bits at i-th positions
int NS = 0;
// Traverse the array
foreach(int j in arr)
{
// If i-th bit of current element is set
if ((getBit & j) != 0)
S++; // Update S
else
NS++; // Update NS
}
// If count of set bits at i-th position is
// greater
// than count of clear bits
if (S > NS)
// Update ans
ans += (int)Math.Pow(2, i);
// Update getBit
getBit <<= 1;
}
Console.Write(ans + " ");
}
// Function to traverse the tree and map all the nodes
// of same vertical line to vertical distance
public static void Trav(Node root, int hd,
Dictionary<int, List<int> > mp)
{
if (root == null)
return;
// Storing the values in the map
if (mp.ContainsKey(hd) == false)
mp[hd] = new List<int>();
mp[hd].Add(root.val);
// Recursive calls on left and right subtree
Trav(root.left, hd - 1, mp);
Trav(root.right, hd + 1, mp);
}
// Function to compress all the nodes on the same
// vertical
// line with a single node that satisfies the condition
public static void compressTree(Node root)
{
// Map all the nodes on the same vertical line
Dictionary<int, List<int> > mp
= new Dictionary<int, List<int> >();
Trav(root, 0, mp);
// Getting the range of horizontal distances
int lower = Int32.MaxValue, upper = Int32.MinValue;
foreach(KeyValuePair<int, List<int> > i in mp)
{
lower = Math.Min(lower, i.Key);
upper = Math.Max(upper, i.Key);
}
for (int i = lower; i <= upper; i++)
evalComp(mp[i]);
}
}
// This code is contributed by Tapesh(tapeshdua420)
Javascript
<script>
// Javascript program for the above approach
// Structure of a node
// of th tree
class TreeNode {
constructor(val = "", left = null, right = null) {
this.val = val;
this.left = left;
this.right = right;
}
}
// Function to compress all the nodes
// on the same vertical line
function evalComp(arr) {
// Stores node by compressing all
// nodes on the current vertical line
let ans = 0;
// Check if i-th bit of current bit
// set or not
let getBit = 1;
// Iterate over the range [0, 31]
for (let i = 0; i < 32; i++) {
// Stores count of set bits
// at i-th positions
let S = 0;
// Stores count of clear bits
// at i-th positions
let NS = 0;
// Traverse the array
for (j of arr) {
// If i-th bit of current element
// is set
if (getBit & j)
// Update S
S += 1;
// Update NS
else NS += 1;
}
// If count of set bits at i-th position
// is greater than count of clear bits
if (S > NS)
// Update ans
ans += 2 ** i;
// Update getBit
getBit <<= 1;
}
document.write(ans + " ");
}
// Function to compress all the nodes on
// the same vertical line with a single node
// that satisfies the condition
function compressTree(root) {
// Map all the nodes on the same vertical line
let mp = new Map();
// Function to traverse the tree and map
// all the nodes of same vertical line
// to vertical distance
function Trav(root, hd) {
if (!root) return;
// Storing the values in the map
if (!mp.has(hd)) mp.set(hd, [root.val]);
else {
let temp = mp.get(hd);
temp.push(root.val);
mp.set(hd, temp);
}
// Recursive calls on left and right subtree
Trav(root.left, hd - 1);
Trav(root.right, hd + 1);
}
Trav(root, 0);
// Getting the range of
// horizontal distances
let lower = [...mp.keys()].sort((a, b) => a - b)[0];
let upper = [...mp.keys()].sort((a, b) => b - a)[0];
for (let i = lower; i <= upper; i++) evalComp(mp.get(i));
}
// Driver Code
let root = new TreeNode(5);
root.left = new TreeNode(3);
root.right = new TreeNode(2);
root.left.left = new TreeNode(1);
root.left.right = new TreeNode(4);
root.right.left = new TreeNode(1);
root.right.right = new TreeNode(2);
// Function Call
compressTree(root);
// This code is contributed by _saurabh_jaiswal.
</script>
Producción:
1 3 5 2 2
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por rohitsingh07052 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA